College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 29

Answer

$x=1.53$

Work Step by Step

Divide both sides by $5$. Then make each side a natural log, and pull down the exponent on the left side. Then divide to solve for $x$. $5e^x=23$ $e^x=\frac{23}{5}$ $x\ln(e)=\ln(\frac{23}{5})$ $x=\frac{\ln(\frac{23}{5})}{\ln(e)}$ $x=1.53$
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